# Chapter 1 Fracture Mechanics

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Chapter 1 Fracture Mechanics, 2nd ed. (2015) Solution Manual
1
CHAPTER 1
THEORY OF ELASTICITY
1.1 A thin sheet made of an aluminum alloy having E = 67 GPa, G = 256 GPa and also v = 1/3
was used for two dimensional surface strain measurements. The measurements provided
10.5 10 , 20 10 ,
5 5
 x   x xx yy   and 240 10 .
5
 x xy  Determine the corresponding stresses.
Solution:

 

MPa
MPa
v
v
v
E
yy xx
xx yy
yy
xx
12.44
2.89
(1 )
2
 
 

MPa
v
E xy
xy 60.45
2(1 )

1.2 Determine (a) the principal stresses and strains and (b) the maximum shear stress for the case
described in Problem 1.1.
Solution:
(a) xy MPa MPa xx yy xx yy 4.28 60.60
2 2
2
2
1,2     

 

 
   

56.32 MPa  1 
64.88 MPa  2  
The principal strains are
5 3
2 2
1,2 4.75 10 1.20 10
2 2 2
 
   

 

 

 x x
xx yy xx yy xy     

3
1
1.15 10
  x
3
2
1.25 10
   x
(b) The maximum
xy MPa xx yy 60.60
2
2
2
max   

 
 
 

Chapter 1 Fracture Mechanics, 2nd ed. (2015) Solution Manual
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1.3 Calculate the diameter of a 1-m long wire that supports a weight of 200 Newton. If the wire
stretches 2 mm, calculate the strain and the stress induced by the weight. Let E = 207 GPa.
Solution:
2 10 0.20%
10
2 3
3
  

x
mm
mm
l
l
o

mm
P
d 0.78 4
 

  E  414 MPa
P A ( / 4)d 200 N
2
     
1.4 Derive an expression for the local uniform strain across the neck of a round bar being loaded
in tension. Then, determine its magnitude if the original diameter is reduced 80%.
Solution:
Volume constancy:
AL  AoLo
   
2
ln L / L ln d / d
L
dL
o o
L
L
zz
o
   

d d  zz o
  2ln /
If 0.80 ,
o
d  d then
 2ln1/ 0.80  0.4463 zz

 zz  44.63%
1.5 The torsion of a bar containing a longitudinal sharp groove may be characterized by a
warping function of the type [after F.A McClintock, Proc. Inter. Conf. on Fracture of Metals,
Inst. of Mechanical Eng., London, (1956) 538] 
  
r
z w ydx xdy
0
  ( ) . The displacements are
 0  x
and  y  rz , where  and r are the angle of twist per unit length and the crack tip
radius, respectively. The polar coordinates have the origin at the tip of the groove, which has a
radius (R). Determine w, the shear strains rz  and .
 z
 In addition, predict the maximum of the
shear strain .
 z

Chapter 1 Fracture Mechanics, 2nd ed. (2015) Solution Manual
3
Solution:
Let x  Rcos and y  Rsin so that

 
 

y z
z
z r
1
dx  Rsind and dy  Rcosd  
r
R
R r
r
r z
2
1 2 
       

 
 
   

   
0
2
0
( ydx xdy) R d
r
z
If r  0 then   z   max  
 0 rz 
1.6 A cantilever beam having a cross-sectional area of 1.5 cm2
is fixed at the left-hand side and
loaded with a 100 Newton downward vertical force at the extreme end as shown in the figure
shown below. Determine the strain in the strain gage located at 8 cm from the fixed end of the
shown steel cantilever beam. The steel modulus of elasticity of is E = 207 GPa.
Solution:
FBD:
Thus,
2
1
3
1
6
12
2
bh
P X X
nh
P( X X )( h / )

 

 
Using Hooke’s law yields the elastic strain
E

 
 
4
2 2 2
2
2
1
4.64 10
/ )(3 10 )(0.50 10 )
6
(207 10
(6)(100 )(20 8 ) 10 6
9

 

 x
x N m x m x m
N m m x
Ebh
P X X

M
X – X1

 
  


1
1
0
0
M P X X
P X X M
M ( Moments)
12
3
2
bh I
h C
I
MC

 
X = 20 cm
b = 3 cm
P = 100 N
h = 0.5 cm
Strain gage – +
8 cm
Chapter 1 Fracture Mechanics, 2nd ed. (2015) Solution Manual
4
1.7 The stress-strain behavior of an annealed low-carbon steel (σys = 200 MPa and E = 207 GPa)
obeys the Hollomon equation with k = 530 MPa and n =0.25. (a) Plot the true and engineering
stress-strain curves. Calculate (b) the tensile strength ( ts ) and (c) the strain energy density up to
the instability point.
Solution:
(a) Plots
 
   
  
 
 
ln 1
1
t
t
 
    
n
n
t t
k
k
  
 
  

1 ln 1
  
 
n
k

1
ln 1
(b) The engineering yield strain and the true strain at the instability point on the true stress-strain
curve are, respectively
-3
3
4.3478×10
207 10
900
  
x MPa
MPa
E
ys
ys

 n  0.25 t
 @ Instability point
At the instability or ultimate tensile strength point (maximum),
  
 
1 exp( ) 1 exp(0.25) 0.28403
ln 1
1,200 0.25 848.53 0.25
max
      
 
  
t
t
n
k t MPa MPa
 
 
 
0 0.05 0.1 0.15 0.2 0.25 0.3
1000
750
500
250
0
0 0.05 0.10 0.15 0.20 0.25 0.30
Strain
Stress
True
Engineering
Yield Point
 ,   0.0043, 305.81 MPa ys  ys 
 0.25 ts 
0
  
1 0.25
1200 ln 1
0.25

 
0.25 1200 t t   
 n  0.25 t

@ Instability point
Chapter 1 Fracture Mechanics, 2nd ed. (2015) Solution Manual
5
 
 
374.77 /1 0.28403 291.87 MPa
/ 1
1
max
max
  
 
 
ts MPa
ts
ts

  
  
(c) The strain energy density (W) is just the area under the stress-strain curve. In general, the true
and engineering strain energy densities are
    
       
3
2
1 0.25 1 0.25 -3
2 -3
1 1
2
0
0
33550 335.50 /
0.28403 4.3478×10 33550
1 0.25
1,200
2
207,000 4.3478×10
2 1
MJ m
m
m
m
MN W .
W . MPa
n
E k W E d k d
W d d
t
t
n
ys
n
ts
n ys
t
Elastic Plastic
t
ts
ys
ys
ts
ys
ys
 
 

 

   

 
 
 
 
 

   
   

and
  
 
  
  
  
W . . i
W
n
E k
d
k W E d
W d d
n
ts ys
ys
n
Elastic Plastic
ts
ys
ys
ts
ys
ys
0 99932 0 95718
ln 0.28403- 4.3478×10
0.25 1
1,200
2
207,000 4.3478×10
ln 1 1
1 2 1
ln 1
0.25 1 -3
2 -3
1
2
0
0
 

 
  

 

  

 
 
 

  
   

Therefore, the engineering strain energy density can not be determined analytically using the
modified Hollomon equation. Instead, the solution can be achieved numerically. Thus,
  
 
3
0.28403
0.0043478
0.0043478 0.25
0
170 60 /
170 60
1
1200 ln 1
207000
W . MJ m
W d d . MPa

   

 
Wt  2W
Chapter 1 Fracture Mechanics, 2nd ed. (2015) Solution Manual
6
1.8 The figure below shows a schematic cross-sectional view of a pressure vessel (hollow
cylinder) subjected to internal and external pressures. Determine the stresses at a point Pr,  in
polar coordinates when (a)  0 Pi
and  0, Pi
(b)  0, Po
(c)  0 Pi
and (d) a  0 so that the
hollow cylinder becomes a solid cylinder. (e) Plot f (r)  rr  and f (r)  rr  . Let a  450 mm,
b  800 mm,  0 Pi
and P  40 MPa. The valid radius range must be 0.45 m  r  0.80 m . Use
the following Airy’s stress function
2
1 2 3   c  c ln r  c r
Along with the boundary conditions
P r b
P r a
rr o
rr i
   
   
0 @
0 @
r
r

 
 
Solution:
(a) Derivatives of 2
1 2 3   c  c ln r  c r
2 3
2
2
2
3
2
2
2
c
r
c
r
c r
r
c
r
  

 

0
0
2
2

0
2

 

r
From eq. (1.58),
0
1 1
2
2
1 1
2
2
2 3
2
2
2
2 3
2
2
2
2

 

  

 

 


r r r
c
r
c
r
c
r
c
r r r
r
rr
Chapter 1 Fracture Mechanics, 2nd ed. (2015) Solution Manual
7
Using the boundary conditions yields
2 3
2
2 3
2
2 3
2
2 3
2
2 2
2 2
c
r
c
c P
r
c
P
c
r
c
c P
r
c
P
rr o o
rr i i
      
      

(a)
Replace r for the proper radius and solve eq. (a) for 2
c and 3
c
2 3
2
2 3
2
2
2
c
r
c
P
c
r
c
P
o
i
  
  
from which
2
2
3
2 3
2
2
2
a
c
c P
c
a
c
P
i
i
  
  
and
2
2
3
2 3
2
2
2
b
c
c P
c
b
c
P
o
o
  
  
Then,
  o i
o i
o i
i o
P P
b a
a b
c
P
P P
b
c
a
c
b
c
P
a
c
P

  
  
2 2
2 2
2
2
2
2
2
2
2
2
2
&
   
2 2
2
2 2
2
3
2 2
2
2 2
2
2 2
2
2 2
2
3
2 2
2 2
2 2
2 2
2 2
2
3
2
2 1
1
2
b a
b P
b a
a P
c
b a
a
P
b a
a P
P
b a
a P
b a
a P
c
P P
b a
a b
P P P
b a
a b
b
P
b
c
c P
i o
o
i
o
i i
o o o i o i o

 

   

     
  Po Pi
b a
a b
c 

 2 2
2 2
2 &
 
2 2
2 2
3
2 b a
a P b P
c
i o

Thus,
 
 
 
 
0
2 2 2
2 2
2 2
2 2
2 2 2
2 2
2 2
2 2



r
i o o i
i o o i
rr
b a r
a b P P
b a
a P b P
b a r
a b P P
b a
a P b P
Chapter 1 Fracture Mechanics, 2nd ed. (2015) Solution Manual
8
(b) If  0, Po
then (c) If  0, Pi
then
0
1
1
2
2
2 2
2
2
2
2 2
2



r
i
i
rr
r
b
b a
a P
r
b
b a
a P
0
1
1
2
2
2 2
2
2
2
2 2
2



r
o
o
rr
r
a
b a
b P
r
a
b a
b P
(d) If a  0, then
 0
 
 



r
o
rr o
P
P
(e) The plot
0.4 0.5 0.6 0.7 0.8
100
75
50
25
0
-25
-50
(MPa)


 rr
Chapter 1 Fracture Mechanics, 2nd ed. (2015) Solution Manual
9
1.9 Consider an infinite plate with a central hole subjected to a remote uniform stress as shown in
Example 1.5. The boundary conditions for this loaded plate are (1) S x  x 
   and σy = τxy = 0
at r   and (2)   0    r r
at r  a. Use the following complex potentials [12]

z 
S
4
1 
2a
2
z
2
and 

z  
S
2
1 
a
2
z
2

3a
4
z
4
to determine σr
, σθ and τrθ (in polar coordinates).
Solution: The infinite plate with a central hole subjected to a remote uniform stress is
Use the following stress expressions
r    2

z  2

z  4 Re

z
  r  i2r  2z
z  

ze
i2
Recall that the Euler’s formula for z and z² are
z  rei  rcos  isin 
z  rei  rcos  isin
z
m  r
me
im  r
3
cosm  isinm
and their real and imaginary parts are
Re z  r cos & Re z
m  r
m cosm
Imz  rsin  & Imz
m  r
m sinm
Apply the Euler’s formula to the above complex potentials so that

z 
S
4
1 
2a
2
z
2

S
4
1 
2a
2
r
2
e
i2

z 
S
4
1 
2a
2
r
2
e
i2
4Re

z  S 1 
2a
2
r
2
cos2
(a) The first stress equation:
r    4 Re
z  S 1 
2a
2
r
2
cos2
Then,
(a)
Chapter 1 Fracture Mechanics, 2nd ed. (2015) Solution Manual
10

z 
S
4

a
2
2z
2


z  
2Sa2
2z
3

Sa2
z
3

Sa2
r
3
e
i3

Sa2
r
3
e
i3


z 
Sa2
r
3
cos3 (Real part)
Thus,
z

z  rei

Sa2
r
3
e
i3

Sa2
r
2
e
i4
(b) The second potential:


z  
S
2
1 
a
2
z
2

3a
4
z
4
 
S
2
1 
a
2
r
2
e
i2

3a
4
r
4
e
i4
Hence,
  r  i2r  2z

z  

ze
i2
  r  i2r  2
Sa2
r
2
e
i4
e
i2 
S
2
e
i2 
Sa2
e
i2
2r
2
e
i2

3Sa4
e
i2
2r
4
e
i4
  r  i2r  S
2a
2
r
2
e
i2  e
i2 
a
2
r
2

3a
4
r
4
e
i2
Real parts:
  r  S
2a
2
r
2
cos2  cos2 
a
2
r
2

3a
4
r
4
cos2
Add eqs. (a) and (c) and solve for σθ
 
S
2
1 
a
2
r
2

S
2
1 
3a
4
r
4
cos2
Use eq. (a) to solve for σr
r 
S
2
1 
a
2
r
2

S
2
1 
4a
2
r
2

3a
4
r
4
cos2
Using the imaginary part of eq. (b) yields shear stress τrθ as
i2r  S i
2a
2
r
2
sin2  isin2  i
3a
4
r
4
sin2
r  
S
2
1 
2a
2
r
2

3a
4
r
4
sin 2
At r = a
r  0
  S  2S cos2  S1  2 cos2
r  0
(b)
(c)
(d1)
(d2)
(d3)
(f1)
(f3)
(f2)
Chapter 1 Fracture Mechanics, 2nd ed. (2015) Solution Manual
11
1.10 Use the Cauchy-Riemann condition to show that (a) f(z) = 1/z is analytic and (b) its
derivative is f(z)= -1/z².
Solution:
(a) If z = x + iy, then
1
z 
1
xiy 
1
xiy
xiy
xiy 
xiy
x
2y
2
1
z 
x
x
2y
2

iy
x
2y
2
Let
u 
x
x
2y
2
v  
y
x
2y
2
Then
u
x

x
2y
2
x
x x
2y
2

x
2y
2
2

x
2y
22x
2
x
2y
2
2

y
2x
2
x
2y
2
2
u
y
 
2xy
x
2y
2
2
and
v
x

2xy
x
2  y
2

2
v
y

x
2  x
2
x
2  y
2

2
Thus,
u
y
 
v
x
Therefore, f(z) = 1/z is analytic because ∂u/∂y = ∂v/∂x.
(b) The derivative of f(z) = 1/z is
f

z 
d
dz 
1
z
 
u
x
 i
v
x
f

z 
y
2x
2
x
2y
2
2
 i
2xy
x
2y
2
2

yix
2
x
2y
2
2

iiyx 
2
i x
2y
2
2

xiy
2
x
2y
2
2
 
z
2
xiyxiy
2
 
z
2
zz
2
 
1
z
2
Chapter 1 Fracture Mechanics, 2nd ed. (2015) Solution Manual
12
1.12 Evaluate the Cauchy integral formula given below for the complex function when zo = π.
fzo  
1
2i

cosz
z
21
dz
Solution:
The integral has to be modified as
fzo  
1
2i

cosz
z
2  1
dz
fzo  
1
2i

cosz
z  1z  1
dz
fzo  
1

fz
z  1
dz

fz
z  1
dz  2ifzo 
where
fz 
cosz
z1
Then,

fz
z  1
dz  2ifzo 

fz
z  1
dz  2i
coszo
zo  1 zo

fz
z  1
dz  2i
cos
  1
 1. 52i
Chapter 2 Fracture Mechanics, 2nd ed. (2015) Solution Manual
1
CHAPTER 2
INTRODUCTION TO FRACTURE
MECHANICS
2.1 Show that the applied stress  0 as the crack tip radius  0. Explain
Solution:
From eq. (2.28),
a
and a 
 

 max  max 2
1
 2 
Therefore,  0 as  0. This means that the applied stress  0 since the material has
fractured or separated, at least into two pieces, and there is not a crack present so that  0. On
the other hand,  0 means that the existing crack is very sharp as in the case of a fatigue
crack.
2.2 For a Griffith crack case, crack propagation takes place if the strain energy satisfies the
inequality U(a) U(a  a) 2a , where a  crack extension and  is the surface energy.
Show that the crack driving force or the strain energy release rate at instability is .
a
U(a)
G

Solution:
Use the Taylor’s series to expand the left side term of the inequality. Thus,
…………..
3
3
( )
3
3!
2 1
2
( )
2
2!
( ) 1
1!
1
( ) ( )  

 

 

    a
a
U a
a
a
U a
a
a
U a
U a a U a (2.1)
Using the first two terms yields
a
a
U a
U a a U a 

   
( )
( ) ( )
(2.2)
a
a
U a
U a U a a 

    
( )
( ) ( ) (2.2)
Chapter 2 Fracture Mechanics, 2nd ed. (2015) Solution Manual
2
Then,
a a
a
U a
  

 2
( )
(2.3)
G
a
U a
 

 2
( )
(2.4)
Therefore,
a
U a G

( )
(2.5)
2.3 A 1mmx15mmx100mm steel strap has a 3-mm long central crack is loaded to failure.
Assume that the steel is brittle and has E  207,000 MPa, 1500 MPa,
ys   and
K 70 MPa m.
IC  Determine the critical stress ( )  c
and the critical strain energy release rate.
Solution:
a = 1.5 mm and a / w  0.10
The geometry correction factor:
( / ) tan   1.07 

  
w
a
a
w
f a w

The stress intensity factor
K a IC    f 
MPa
a
KIC
c
1020
2
 
 

The strain energy release rate
E
KIC
IC
2
G 
GIC  23.70 kPa m
2 GIC  23.70 kN / m
2.4 Suppose that a structure made of plates has one cracked plate. If the crack reaches a
critical size, will that plate fracture or the entire structure collapse? Explain.
Answer: If the structure does not have crack stoppers, the entire structure will collapse since the
cracked plate will be the source of structural instability.
Chapter 2 Fracture Mechanics, 2nd ed. (2015) Solution Manual
3
2.5 What is crack instability based on according to Griffith criterion?
Answer: It is based on the energy consumption creating new surfaces or in developing crack
extension. This energy is the surface energy 2.
2.6 Can the Griffith Theory be applied for a quenched steel containing 1.2%C, if a pennyshaped crack is detected?
Answer: Despite the quenched steel is brittle, the Griffith Theory applies only if a plastic zone
does not form at the crack tip.
2.7 Will the Irwin Theory (or modified Griffith Theory) be valid for a changing plastic zone size
Answer: No. It is valid if the plastic zone remains constant as it the case in many materials
under plane strain conditions.
2.8 What are the major roles of the surface energy and the stored elastic energies in a crack
growth situation?
Answer: The surface energy acts as a retarding force for crack growth and the stored elastic
energy acts to extend the crack. Thus, concurrent with crack growth is the recovery of elastic
strain energy by the relaxation of atomic bonds above and below the fracture plane.
2.9 What does happen to the elastic strain energy when crack growth occur?
Answer: It is released as the crack driving force for crack growth.
2.10 What does U/a = 0 mean?
Answer: Crack growth occurs at ,   f
 as shown in eq. (2.23), since the magnitude of crack
growth and the crack resistance force become equal. That is,
E
a
s
2
2
 
 
If
E
a
s
2
2
 
  , then fracture occurs and the fracture toughness without any plastic zone
deformation is
s
γ
c
G  2 at ,   f
 as shown in eq. (2.31) with p  0. Thus,
s
γ
c
G  2 is related
to an irreversible process of fracture.
Chapter 2 Fracture Mechanics, 2nd ed. (2015) Solution Manual
4
2.11 Derive eq. (2.28) starting with eq. (2.20).
Solution:
From eq. (2.20),
Multiplying this expression by  yields the stress intensity factor as per
eq. (2.28)
a
a
   

    max 2
1
K a I
  

 

a
a
Kt

 
max
max
2
1
2
Chapter 3 Fracture Mechanics, 2nd ed. (2015) Solution Manual
1
CHAPTER 3
LINEAR ELASTIC FRACTURE
MECHANICS
3.1 A steel strap 1-mm thick and 20-mm wide with a through-the-thickness central crack 4-mm
long is loaded to failure. (a) Determine the critical load if KIC = 80 MPa m for the strap
material. (b) Use the available correction factor,   f (a / w), for this crack configuration. Now,
do the following three comparisons to have a better understanding: Feddersen:
  c
 ( fraction) , Irwin:   c
 ( fraction) and Koiter-Benthem:   c
 ( fraction) .
Solution:
B = 1mm, w = 20mm, 2a = 4 mm, a/w = 0.10, KIC = 80 MPa m , K a f a w I c
    / ,
(a) Finite plate:
From eqs. (3.31) and (3.32) along with o o
a / w  0.10 rad.  (180 )(0.10) /  5.7296
Feddersen [12]: f    
a
w
a
w
 sec  .
 1025  c
 984 64 . MPa
Irwin [13]: f  
a
w
w
a
a
w



tan 1017 .  c
 992 38 . MPa
Koiter-Benthem [14]: f      
a
w a
w
a
w
a
w

 

1
1 2
1 1304 1021
2
. .  c
 988 49 . MPa
(b) Infinite plate approach [a/w  0 since f(a/w)  1].
 
MPa
m
MPa m
a
KIC
c 1009.25
2 10
80
3

 
  

Feddersen: 97.56% (infinite)  c   c
Irwin: 98.33% (infinite)  c   c
Koiter: 97.94% (infinite)  c   c
Therefore, the calculated  c values do not differ much since the plate dimensions are large
enough and the initial crack size is relatively small.
B=1 mm
P 2a w P
Chapter 3 Fracture Mechanics, 2nd ed. (2015) Solution Manual
2
3.2 A steel tension bar 8-mm thick and 50-mm wide with an initial single-edge crack of 10-mm
long is subjected to a uniaxial stress  = 140 MPa. (a) Determine the stress intensity factor KI
. If
KIC = 60 MPa m , is the crack stable? (b) Determine the critical crack size, and (c) determine the
Solution:
 = P/A= 140 MPa, KIC = 60 MPa m , 

  
w
a
K a f
I  
B = 8 mm, w = 50 mm, a = 10 mm, x = a/w = 0.20, A = Bw
From Table 3.1, KI
   a and
         
2 3 4
  f x  1.12  0.231 x  10.55 x  21.71 x  30.38 x
  f x  0.20  1.37
(a) The applied stress intensity factor
KI = 34 MPa m
KIC = 60 MPa m
Therefore, the crack is stable because KI 5 mm (Not valid)
c) B  7 mm
3.4 (a) One guitar steel string has a miniature circumferential crack of 0.009 mm deep. This
implies that the radius ratio is almost unity, d / D  1. b) Another string has a localized miniature
surface crack (single-edge crack like) of 0.009 mm deep. Assume that both
strings are identical with an outer diameter of 0.28 mm. If a load of 49 N is
applied to the string when being tuned, will it break? Given properties:
K 15MPa m, 795MPa.
IC ys   
Solution:
(a) Circumferential crack
If mm
D d
a 0.009
2

 , then d  0.262 mm
MPa
x m
N
d
P
908.87
(0.262 10 )
4 (4)(49 )
2 3
  

 

 



    
2 3
0.36 0.73
8
3
2
1
2
1
( / )
D
d
D
d
D
d
d
D
d
D
 f d D
For d / D  0.9357,
  1.14

Chapter 3 Fracture Mechanics, 2nd ed. (2015) Solution Manual
4
K  MPa  m
K a
I
I
3
(1.14) 908.87 0.009 10
 

  
KI
 5.51 MPa m
(b) Surface edge crack
  1.12
KI
   a
K  MPa  m I
3
(1.12) 908.87 0.009 10
  
KI
 5.41 MPa m
Therefore, the strings will not break since KI  KIC in both cases. When d / D 1 and the crack
size is very small, either approach gives similar results.
3.5 A 7075-T6 aluminum alloy is loaded in tension. Initially the 10mmx100mmx500mm plate
has a 4-mm single-edge through–the-thickness crack. (a) Is this test valid? (b) Calculate the
maximum allowable tension stress this plate can support, (c) Is it necessary to correct KI due to
crack-tip plasticity? Why? or Why not? (d) Calculate the design stress and stress intensity factor
if the safety factor is 1.5. Data: ys = 586 MPa and KIC = 33 MPa m .
Solution:
(a) mm
K
B
ys
IC 2.5 7.93
2
 

 

and  0.0375
w
a
Plane strain condition holds because the actual thickness is greater than the required value.
(b) KI
   a ; a = 4 mm and w = 100 mm
 
2 3 4
1.12 0.231 10.55 21.71 30.38 

 

 



   
w
a
w
a
w
a
w
a
w
a  f
   f 0.04  1.13
w
a  f
Thus,
MPa
a
KIC
  261
 

Chapter 3 Fracture Mechanics, 2nd ed. (2015) Solution Manual
5
(c) Plastic zone correction:
mm
K
r
ys
IC 0.17
6
1
2
 

 
 
K a a a r eff     eff with eff   = 4.17 mm
    
K MPa m
K MPa x m
eff
eff
33.76
1.13 261 4.17 10 3

Therefore, there is no need for crack tip plasticity since KI,eff  KIC .
(d) MPa MPa
SF d 174
1.5
261
  

MPa m
MPa m
SF
K
K
IC
Id 22
1.5
33
  
3.6 A steel ship deck (30mm thick, 12m wide, and 20m long) is stressed in the manner shown
below. It is operated below its ductile-to-brittle transition temperature (with KIc = 28.3 MPa m ).
If a 65-mm long through–the-thickness central crack is present, calculate the tensile stress for
catastrophic failure. Compare this stress with the yield strength of 240 MPa for this steel.
Solution:
2a = 65mm   f a / w 1 since a/w 0
w = 12 m
L = 20m
P 2a P
T
KI
(MPa m )
or
CVN
Energy
(Joules)
Transition Range.
Chapter 3 Fracture Mechanics, 2nd ed. (2015) Solution Manual
6
K a IC     KIC  28.3 MPa m and  ys  240MPa
MPa
m
MPa m
a
KIC 88.57
10
2
65
28.3
3

 
 

Therefore,    ys and to assure structural integrity, the safety factor (SF) that can be used to
avoid fracture or crack propagation is in the order of
2.71
88.57
240
  
MPa
MPa SF ys

3.7 Show that  0
da
KI
for crack instability in a large plate under a remote tensile external stress.
Solution:
K a I
  
0
2
1
 
da a
dKI 
 since   0 and a  0
3.8 The plate below has an internal crack subjected to a pressure P on the crack surface. The
stress intensity factors at points A and B are
 

 dx
a x
a x
a
P
KA

 

 dx
a x
a x
a
P
KB

Use the principle of superposition to show that the total stress intensity factor is of the form
KI
 P a
Solution:
According to the principle of superposition, the total stress intensity factor is
 

  

  
2 2
2
a x
a dx dx P
a x
a x
a x
a x
a
P
KI KA KB
 
P
A
B
2a
Chapter 3 Fracture Mechanics, 2nd ed. (2015) Solution Manual
7
Furthermore,
2 2
2
a x
a
( a x )( a x )
( a x ) ( a x )
a x
a x
a x
a x
a x
a x
a x
a x

 
  

 
 a
x
ar
a x
dx
csin
2 2
and
a
a x
KI 2P arcsin

If x = a on the crack surface, then  
o
arcsin 1   / 2  90 and
a P a
P
a
x
ar
a
KI P 

 
   
2
2
2 sin
3.9 A pressure vessel is to be designed using the leak-before-break criterion based on the
circumferential wall stress and plane strain fracture toughness. The design stress is restricted by
the yield strength  ys and a safety factor (SF). Derive expressions for (a) the critical crack size
and (b) the maximum allowable pressure when the crack size is equals to the vessel thickness.
Solution:
Select some alloy having known KIC and  ys values. The alloy with the highest
ys
KIC

2
and
2

ys
KIC

values should be used for constructing the pressure vessel.
(a) The critical crack size
F
ys
d
IC c
S
K a

  

Thus,
2
2
1

 

ys
F IC
c
c
F
ys
IC
S K
a
a
S
K
  

 
Chapter 3 Fracture Mechanics, 2nd ed. (2015) Solution Manual
8
3.10 A stock of steel plates with GIC = 130 kJ/m², σys = 2,200 MPa, E = 207 GPa, v = 1/3 are
used to fabricate a cylindrical pressure vessel (di = 5 m and B = 25.4 mm). The vessel fractured
at a pressure of 20 MPa. Subsequent failure analysis revealed an internal semi-elliptical surface
crack of a = 2.5 mm and 2c = 10 mm. (a) Use a fracture mechanics approach to predict the
critical crack length this steel would tolerate. (b) Based on this catastrophic failure, another
vessel was constructed with do = 5.5 m and di = 5 m. Will this new vessel fracture at a pressure
of 20 MPa if there is an internal semi-elliptical surface crack having the same dimensions as in
part (a)?
Solution:
(a) The vessel is based on the thin-wall theory since the thickness is
mm
d x mm B
i 250
20
5 10
20
3
   (Requirement)
B  25.4 mm (Given) is less than the required thickness. So use the thin-wall theory.
Thus, the fracture hoop stress is
h 
Pidi
2B

20 MPa 5×103 mm
2 25.4 mm
h  1,968.50 MPa
Using eq. (2.34) yields the critical crack length
ac 
EGIC
 1  v
2
h
2
ac 
207×103 MPa 130×103 MPa.m
 1  1/32
 1,968.50 MPa 2
ac  2.49 mm  2.5 mm
(b) For the new vessel,
B 
do  di
2

5.6 m  5 m
2
 0.30 m
do
di

5.6
5
 1.12
Also,
h 
do/di
2
 1
do/di
2
 1
Pi  8.86Pi
h  8.8620 MPa  177.23 MPa
  h  Pi  9.86Pi  197.23 MPa
Chapter 3 Fracture Mechanics, 2nd ed. (2015) Solution Manual
9
Then,
a/2c  2.5/10  0.25
/ys  197.23/2,200  0.09
From eq. (3.44),
Q 

2
2
3
4

a
2c
2 2

7
33

ys
2
Q 

2
2
3
4
 0.25
2
2

7
33 0.09
2
Q  1.63
Thus, the applied stress intensity factor is
KI  
 a
Q
KI  197.23 MPa  2.5×103

1.63
KI  13.69 MPa m
From eq. (2.33) at fracture,
KIC 
EGIC
1  v
2
KIC 
207×103 MPa 130×103 MPa.m
1  1/3
2
KIC  174 MPa m
Therefore, the new vessel will not fracture because KI < KIC. 3.11 A cylindrical pressure vessel with B = 25.4 mm and di = 800 mm is subjected to an internal pressure Pi . The material has KIC  31 MPa m and σys = 600 MPa. (a) Use a safety factor to determine the actual pressure Pi . (b) Assume there exists a semi-elliptical surface crack with a = 5 mm and 2c = 25 mm and that a pressure surge occurs causing fracture of the vessel. Calculate the fracture internal pressure Pf . (c) Calculate the critical crack length. Solution: (a) Let's figure out which pressure vessel wall theory should be used. B  di 20  800 mm 20  40 mm Chapter 3 Fracture Mechanics, 2nd ed. (2015) Solution Manual 10 Therefore, the thin-wall theory should be used. The actual or applied pressure is h  Pidi 2B  ys SF Pi  2Bys SFdi  2 25.4 mm 600 MPa 2.5 800 mm Pi  15.24 MPa (b) Using the principle of superposition yields   h  Pi  Pidi 2B  Pi   di 2B  1 Pi  800 2×25.4  1 Pi   16.75Pi  16.75 15.24 MPa   255.24 MPa (Actua l) Also, a 2c  5 25  0.20  ys  255.24 600  0.43 a B  5 25.4  0.20 M  Mk  1 [See eq.(3.46) & Figure 3.6] From eq. (3.44), Q   2 2 3 4  a 2c 2 2  7 33  ys 2 Q   2 2 3 4  0.2 2 2  7 33 0.43 2 Q  1.50 From eq. (3.41), KIC  MMk  a Q  16.75Pf  a Q Now, the surge pressure is Pf  KIC 16.75 Q  a  31 MPa m 16.75 1.50  5×103 m Pf  18.09 MPa Chapter 3 Fracture Mechanics, 2nd ed. (2015) Solution Manual 11 (c) The critical crack length is KIC    a Q  16.75Pi  ac Q ac  Q  KIC 16.75Pi 2  1.5  31 MPa m 16.75  15.24 MPa 2 ac  7.04 mm 3.12 This is a problem that involves strength of materials and fracture mechanics. An AISI 4340 steel is used to design a cylindrical pressure having an inside diameter and an outside diameter of 6.35 cm and 12.07 cm, respectively. The hoop stress is not to exceed 80% of the yield strength of the material. (a) Is the structure a thin-wall vessel or a thick-wall pipe? (b) What is the internal pressure? (c) Assume that an internal semi-elliptical surface crack exist with a = 2 mm and 2c = 6 mm. Will the vessel fail? (d) Will you recommend steel for the pressure vessel? Why? or Why not? (e) What is the maximum crack length the AISI 4340 steel can tolerate? Explain. Solution: Internal Crack External Surface From Table 3.2, σys = 1,476 MPa and KIC = 81 MPa for AISI 4340 steel. (a) Using the diameters yields do di  12.07 6.35  1.09 B  do  di 2  12.07 cm  6.35 cm 2 B  2.86 cm Therefore, the pressure vessel is based on the thick-wall theory because do di  1.1 B  di 20  0.32 (b) From eq. (3.43e), h  do/di 2  1 do/di 2  1 Pi  0.8ys Pi  0.8ys do/di 2  1 do/di 2  1 di  6.35 m dO  12.07 m a 2c a  2 mm and 2c  6 mm Chapter 3 Fracture Mechanics, 2nd ed. (2015) Solution Manual 12 Thus, the internal pressure is Pi  0.8 1,476 MPa 1.09 2  1 1.09 2  1  Pi  101.51 MPa (c) Fracture mechanics:   h  Pi  0.8ys  Pi   0.81,476 MPa  101.51 MPa   1,282.30 MPa Thus, a/2c = 2/6 = 0.33, σ/σys = 0.87 and the shape factor becomes Q   2 2 3 4  a 2c 2 2  7 33  ys 2 Q   2 2 3 4  0.33 2 2  7 33 0.87 2 Q  1.66 Then, KI    a Q  1,282.30 MPa  2×103 m 1.66 KI  78.89 MPa m Therefore, the pressure vessel will not fail because KI > B = 15 mm, di = 2 m) is to be made out of a weldable steel alloy
having σys = 1,200 MPa and KIC = 85 MPa. If an embedded elliptical crack (2a = 5 mm and 2c =
16 mm) as shown below is perpendicular to the hoop stress, due to welding defects, the given
data correspond to the operating room temperature and the operating pressure is 8 MPa, then
calculate the applied stress intensity factor. Will the pressure vessel explode?
Solution:
do  di  2B  2 m  2  15  103 m  2.03 m
do
di

2.03
2
 1.02
B 
di
20 
2  103 mm
20  100 mm
The pressure vessel is of thin-wall type because do/di  1.1 and B  di/20. The hoop stress and
the stress intensity factor for the embedded elliptical crack are, respectively
where
Then,
The pressure vessel will not explode because KI a and x a, these stress functions become
x  Re Z 
S
2
z
z
2  a
2
 1 
S
2
x
x
2  a
2
 1
y  Re Z 
S
2
x
x
2  a
2
 1
xy  0
#
For large z = x + iy = x since y = 0, the elastic stresses far from the crack tip vanish when x >> a
along the x-axis. Therefore,
x  0
y  0
xy  0
4.8 Consider the elliptical crack shown below and assume that the crack is in the z-plane where
z    a and pz  p . Derive the stress equations using the given crack geometry and the
Westergaard complex method.
pz  p 
ζ
z
θ
a
x
iy
Solution:
Let us move the origin to crack tip. Then the coordinates of P can be expressed in complex
variable z    a . Use the Westergaard complex equation so that
Zz 

1  a/z
2

z
z
2  a
2
Z 
  a
  2a

a 1 

a
2a 1 

2a
Chapter 4 Fracture Mechanics, 2nd ed. (2015) Solution Manual
9

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Chapter 1 Fracture Mechanics was first posted on October 16, 2019 at 7:33 am.